Isosceles triangle ABC, with AB = BC, is inscribed in a circle with center O.If angle OBC = 10°, what is the measure of angle ACO?A) 10°B) 70°C) 80°D) 160°

1. Break Down the Problem

• We are given an isosceles triangle $ABC$ with $AB = BC$.
• The circle contains this triangle, and the center of the circle is point $O$.
• We know that $\angle OBC = 10^\circ$ and need to find $\angle ACO$.

2. Relevant Concepts

• In triangle $ABC$, since $AB = BC$, we have $\angle ABC = \angle ACB$.
• The angles at the center of a circle and the inscribed angle subtended by the same arc are related by a factor of 2: the inscribed angle is half the central angle.
• Therefore, if we can express $\angle ABC$ using $\angle OBC$, we should be able to find the required angle.

3. Analysis and Detail

• Let $\angle ABC = \angle ACB = x$.
• Since $O$ is the center of the circle, $OB$ and $OC$ are radii, making triangle $OBC$ isosceles with $OB = OC$ (radii of the circle).
• The internal angles in triangle $OBC$ give us $\angle OBC + \angle OCB + \angle BOC = 180^\circ$.
• Since $\angle OBC = 10^\circ$ and $\angle OCB = 10^\circ$, we have: $10^\circ + 10^\circ + \angle BOC = 180^\circ$ $\angle BOC = 180^\circ - 20^\circ = 160^\circ$
• The angle $\angle BAC$ is the external angle for triangle $OBC$: $\angle BAC = \angle BOC = 160^\circ$
• Since triangle $ABC$ is isosceles, $\angle BAC + 2x = 180^\circ$ $160^\circ + 2x = 180^\circ$ $2x = 180^\circ - 160^\circ = 20^\circ \implies x = 10^\circ$
• Therefore, $\angle ABC = \angle ACB = 10^\circ$.

4. Verify and Summarize

• Now we need to find $\angle ACO$:
  • Knowing that $O$ is on the bisector of $\angle ACB$ (as AO = BO = OC), we have: $\angle ACO = \frac{1}{2} \times \angle BAC = \frac{1}{2} \times 160^\circ = 80^\circ$

Final Answer

The measure of angle $ACO$ is $\boxed{80^\circ}$.